A progressive wave of frequency 500 Hz is traveling with a speed of 350 m/s. A compression maximum appears, at a place at a given instant. The minimum time interval after which a rarefaction maximum occurs, at the same place is:
Correct answer: C. 1/1000 s
- A. 1/250 s
- B. 1/500 s
- C. 1/1000 s
- D. 1/350 s
Explanation
The wave speed (v) is given as 350 m/s, and the frequency (f) of the wave is 500 Hz, which means the time period (T) is T = 1/f = 1/500 s.The wavelength (λ) of the wave is related to the wave speed and the time period as follows: λ = v x T.Substitute the values: λ = 350 m/s x 1/500 s = 0.7 m.The time interval for a rarefaction maximum to occur at the same place is equal to the time taken for one full wavelength (from compression maximum to rarefaction maximum) to pass by.Time interval = T/2 = 1/2×500s = 1/1000. So, the correct option is C) 1/1000 s.
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Progressive waves transfer energy through oscillations, with wavelength, frequency, amplitude and wave speed related by v = fλ; sound speed depends on the medium. Superposition produces interference and stationary waves in strings and organ pipes, while simple harmonic motion describes the oscillation itself and must be distinguished from wave propagation.
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